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I am very confused with the signs of standard reduction potentials. For example the standard reduction potential of Zinc is −0.76 V and that of copper is +0.34 V.

I cannot understand the significance of the signs.

As far as I know,if the value is negative then the reaction is non-spontaneous(from the relation between Gibbs Free Energy change and Electrode potential).But if the reaction is non-spontaneous, then why are we classifying it as an electrochemical cell?

Should'nt it be classified as an electrolytic cell? (Sorry if that sounded incorrect but my basics are weak)

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  • $\begingroup$ What reaction are you talking about and why is that important? Also, if the reaction isn't spontaneous, then the reverse reaction is, so the cell will still produce some current, only in another direction. Why wouldn't we call it an electrochemical cell? $\endgroup$ Oct 4, 2015 at 18:04
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    $\begingroup$ @IvanNeretin I am talking about a galvanic cell.I am not talking about any reaction in specific.I am just asking about the signs of the Electrode potentials. $\endgroup$ Oct 4, 2015 at 18:17
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    $\begingroup$ Well, then your understanding is about right: the sign of potential tells us whether certain reaction will be spontaneous or not. What's confusing about that? $\endgroup$ Oct 4, 2015 at 18:27

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There are several questions here so let me start by defining the different types of cells and standard reduction potentials.

Definition of Electrochemical (Galvanic) cell: A cell that converts chemical energy into electrical energy. In these cells a redox reaction creates electrons that can do work.

Definition of Electrolytic cell: A cell that converts electrical energy into chemical energy. In these cells the electrical energy source provides the electrons to perform a reaction.

Standard reduction potential: The tendency for a chemical species to be reduced and is measured in volts at STP. The more positive the potential is the more likely it will be reduced.

Now, here are your two reduction equations for $\ce{Zn}$ and $\ce{Cu}$:

$$\begin{alignat}{2} \ce{Zn^2+(aq) + 2e- &-> Zn}\qquad&{-}0.76\ \mathrm V\\ \ce{Cu^2+(aq) + 2e- &-> Cu}\qquad&{+}0.34\ \mathrm V \end{alignat}$$

The more positive the potential the more favorable the reaction as it is written will be. Remember that $\Delta G = -nFE^\circ$ and that when $\Delta G$ is positive, the reaction is non-spontaneous, and when $\Delta G$ is negative, the reaction is spontaneous. Positive values of $E^\circ$ will lead to negative values of $\Delta G$ and vice versa.

So, the reduction of $\ce{Cu^2+}$ to form $\ce{Cu}$ is more favorable than the reduction of $\ce{Zn^2+}$ to form $\ce{Zn}$. This means that $\ce{Cu^2+}$ is a better oxidant than $\ce{Zn^2+}$. For an electrochemical cell, the cell potential can be calculated by the following equation:

$$E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}$$

For a working electrochemical cell we need $E^\circ_\text{cell}$ to be positive. I would use $\ce{Zn}$ as the anode (oxidation) and $\ce{Cu^2+}$ as the cathode to give an $E^\circ_\text{cell}$ of $+1.10\ \mathrm V$.

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  • $\begingroup$ thanks for the edit Ivan wasn't sure how the arrows worked $\endgroup$
    – Aaron
    Oct 4, 2015 at 22:11

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